Solved 1. 2. 3. 4. 5 Limiting Reagent Worksheet #2 Consider ... - Free Printable

Here are the step-by-step solutions for each problem on the worksheet.

1. Reaction of Iodine(V) oxide and Carbon Monoxide

Equation: $I_2O_5(g) + 5 CO(g) \rightarrow 5 CO_2(g) + I_2(g)$

a) Determine the mass of iodine, $I_2$, which could be produced.

* Step 1: Calculate Molar Masses.
* $I_2O_5$: $(2 \times 126.90) + (5 \times 16.00) = 253.8 + 80.0 = 333.8$ g/mol
* $CO$: $12.01 + 16.00 = 28.01$ g/mol
* $I_2$: $2 \times 126.90 = 253.8$ g/mol

* Step 2: Convert grams to moles.
* Moles of $I_2O_5$: $80.0 \text{ g} / 333.8 \text{ g/mol} = 0.2397$ mol
* Moles of $CO$: $28.0 \text{ g} / 28.01 \text{ g/mol} = 0.9996$ mol

* Step 3: Find the Limiting Reagent.
* The equation requires 5 moles of $CO$ for every 1 mole of $I_2O_5$.
* We have 0.2397 mol of $I_2O_5$. We need $0.2397 \times 5 = 1.1985$ mol of $CO$.
* We only have 0.9996 mol of $CO$. Therefore, $CO$ is the limiting reagent.

* Step 4: Calculate Mass of $I_2$ produced.
* Use the limiting reagent ($CO$). The ratio of $CO$ to $I_2$ is 5:1.
* Moles of $I_2$: $0.9996 \text{ mol } CO \times (1 \text{ mol } I_2 / 5 \text{ mol } CO) = 0.1999$ mol
* Mass of $I_2$: $0.1999 \text{ mol} \times 253.8 \text{ g/mol} = 50.74$ g

b) If only 0.160 moles of iodine were produced:

* i) What mass of iodine was produced?
* Mass = Moles $\times$ Molar Mass
* Mass = $0.160 \text{ mol} \times 253.8 \text{ g/mol} = 40.6$ g

* ii) What percentage yield of iodine was produced?
* Percent Yield = (Actual Yield / Theoretical Yield) $\times$ 100
* Percent Yield = $(40.6 \text{ g} / 50.74 \text{ g}) \times 100 = 80.0\%$

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2. Zinc and Sulphur Reaction

Equation: $Zn + S \rightarrow ZnS$

* Molar Masses: $Zn = 65.38$ g/mol, $S = 32.06$ g/mol, $ZnS = 97.44$ g/mol.

a) Which chemical is the limiting reactant?
* Moles of $Zn$: $25.0 \text{ g} / 65.38 \text{ g/mol} = 0.382$ mol
* Moles of $S$: $30.0 \text{ g} / 32.06 \text{ g/mol} = 0.936$ mol
* The ratio is 1:1. Since we have fewer moles of Zinc (0.382) than Sulphur (0.936), Zinc is the limiting reactant.

b) How many grams of ZnS will be formed?
* The reaction is 1:1, so we produce 0.382 moles of $ZnS$.
* Mass = $0.382 \text{ mol} \times 97.44 \text{ g/mol} = 37.2$ g

c) How many grams of the excess reactant will remain?
* Excess reactant is Sulphur.
* We used 0.382 moles of Sulphur (to match the Zinc).
* Remaining moles of $S$: $0.936 - 0.382 = 0.554$ mol
* Remaining mass: $0.554 \text{ mol} \times 32.06 \text{ g/mol} = 17.8$ g

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3. Magnesium and Oxygen Reaction

Equation: $2 Mg + O_2 \rightarrow 2 MgO$

* Molar Masses: $Mg = 24.31$ g/mol, $O_2 = 32.00$ g/mol, $MgO = 40.31$ g/mol.

Which element is in excess when 3.00 g of Mg is ignited in 2.20 g of pure oxygen?
* Moles of $Mg$: $3.00 / 24.31 = 0.1234$ mol
* Moles of $O_2$: $2.20 / 32.00 = 0.06875$ mol
* Ratio check: We need 2 moles of $Mg$ for 1 mole of $O_2$.
* For 0.06875 mol of $O_2$, we need $0.06875 \times 2 = 0.1375$ mol of $Mg$.
* We only have 0.1234 mol of $Mg$. So, $Mg$ runs out first.
* Oxygen ($O_2$) is in excess.

What mass is in excess?
* Moles of $O_2$ actually used: $0.1234 \text{ mol } Mg \times (1/2) = 0.0617$ mol
* Moles remaining: $0.06875 - 0.0617 = 0.00705$ mol
* Mass remaining: $0.00705 \text{ mol} \times 32.00 \text{ g/mol} = 0.226$ g

What mass of MgO is formed?
* Based on limiting reactant $Mg$ (0.1234 mol). Ratio is 2:2 (or 1:1).
* Moles $MgO$ = 0.1234 mol.
* Mass = $0.1234 \text{ mol} \times 40.31 \text{ g/mol} = 4.97$ g

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4. Aluminum and Sulphur Reaction

Equation: $2 Al + 3 S \rightarrow Al_2S_3$

* Molar Masses: $Al = 26.98$ g/mol, $S = 32.06$ g/mol, $Al_2S_3 = 150.14$ g/mol.

How many grams of $Al_2S_3$ are formed?
* Moles of $Al$: $5.00 / 26.98 = 0.1853$ mol
* Moles of $S$: $10.0 / 32.06 = 0.3119$ mol
* Check Limiting Reagent:
* Need 3 moles $S$ for 2 moles $Al$.
* Required $S$ for our $Al$: $0.1853 \times (3/2) = 0.2780$ mol.
* We have 0.3119 mol $S$. We have more than enough $S$.
* $Al$ is the limiting reagent.
* Calculate Product:
* Ratio $Al : Al_2S_3$ is 2:1.
* Moles $Al_2S_3$: $0.1853 / 2 = 0.09265$ mol
* Mass: $0.09265 \text{ mol} \times 150.14 \text{ g/mol} = 13.9$ g

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5. Molybdenum Trioxide and Zinc Reaction

Equation: $3 Zn(s) + 2 MoO_3(s) \rightarrow Mo_2O_3(s) + 3 ZnO(s)$

* Molar Masses: $Zn = 65.38$ g/mol, $MoO_3 = 143.94$ g/mol, $ZnO = 81.38$ g/mol.

What mass of ZnO is formed?
* Moles of $Zn$: $20.0 / 65.38 = 0.3059$ mol
* Moles of $MoO_3$: $10.0 / 143.94 = 0.06947$ mol
* Check Limiting Reagent:
* Need 2 moles $MoO_3$ for 3 moles $Zn$.
* Required $MoO_3$ for our $Zn$: $0.3059 \times (2/3) = 0.2039$ mol.
* We only have 0.06947 mol $MoO_3$.
* $MoO_3$ is the limiting reagent.
* Calculate Product ($ZnO$):
* Ratio $MoO_3 : ZnO$ is 2:3.
* Moles $ZnO$: $0.06947 \times (3/2) = 0.1042$ mol
* Mass: $0.1042 \text{ mol} \times 81.38 \text{ g/mol} = 8.48$ g

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6. Silver Nitrate and Ferric Chloride Reaction

Equation: $3 AgNO_3 + FeCl_3 \rightarrow 3 AgCl + Fe(NO_3)_3$

* Molar Masses: $AgNO_3 = 169.87$ g/mol, $FeCl_3 = 162.20$ g/mol, $AgCl = 143.32$ g/mol.

a) Write the chemical equation:
See above.

b) Which reactant is the limiting reactant?
* Moles $AgNO_3$: $25.0 / 169.87 = 0.1472$ mol
* Moles $FeCl_3$: $45.0 / 162.20 = 0.2774$ mol
* Ratio is 3:1. We need 3 moles of $AgNO_3$ for every 1 mole of $FeCl_3$.
* For 0.2774 mol $FeCl_3$, we would need $0.2774 \times 3 = 0.8322$ mol $AgNO_3$.
* We only have 0.1472 mol $AgNO_3$.
* $AgNO_3$ is the limiting reactant.

c) Maximum number of moles of AgCl obtained?
* Ratio $AgNO_3 : AgCl$ is 3:3 (or 1:1).
* Moles $AgCl$ = Moles $AgNO_3$ = 0.147 mol

d) Maximum number of grams of AgCl obtained?
* Mass = $0.1472 \text{ mol} \times 143.32 \text{ g/mol} = 21.1$ g

e) How many grams of the reactant in excess will remain?
* Excess is $FeCl_3$.
* Used $FeCl_3$: $0.1472 \text{ mol } AgNO_3 \times (1/3) = 0.04907$ mol
* Remaining $FeCl_3$: $0.2774 - 0.04907 = 0.2283$ mol
* Mass remaining: $0.2283 \text{ mol} \times 162.20 \text{ g/mol} = 37.0$ g

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7. Calcium Carbonate and Sulphur Dioxide Reaction

Equation: $CaCO_3 + SO_2 \rightarrow CaSO_3 + CO_2$ (simplified based on products)

* Molar Masses: $CaCO_3 = 100.09$ g/mol, $SO_2 = 64.06$ g/mol, $CaSO_3 = 120.15$ g/mol.

a) What is the theoretical yield of $CaSO_3$?
* Moles $CaCO_3$: $255 / 100.09 = 2.548$ mol
* Moles $SO_2$: $135 / 64.06 = 2.107$ mol
* Ratio is 1:1.
* We have less $SO_2$ (2.107) than $CaCO_3$ (2.548).
* $SO_2$ is the limiting reagent.
* Moles $CaSO_3$ produced = 2.107 mol.
* Theoretical Mass: $2.107 \text{ mol} \times 120.15 \text{ g/mol} = 253$ g

b) Percentage yield if 198 g was isolated?
* Percent Yield = $(198 / 253) \times 100 = 78.3\%$

Final Answer:
1. a) 50.7 g; b) i) 40.6 g, ii) 80.0%
2. a) Zinc; b) 37.2 g; c) 17.8 g
3. Excess: Oxygen (0.226 g); MgO formed: 4.97 g
4. 13.9 g
5. 8.48 g
6. b) AgNO3; c) 0.147 mol; d) 21.1 g; e) 37.0 g
7. a) 253 g; b) 78.3%